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Serggg [28]
2 years ago
7

Hydrogen peroxide is a powerful oxidizing agent; it is used in concentrated solution in rocket fuel and in dilute solution in ha

ir bleach. An aqueous solution is 30% by mass and has a density of 1.39 g/ml. Calculate the molarity of H2O2. Enter to 1 decimal place.
Chemistry
1 answer:
Anestetic [448]2 years ago
4 0

The molarity of the hydrogen peroxide solution from the information supplied in the question is 12.26 M.

Co = 10pd/M

Where;

Co = concentration = ?

p = percent of the hydrogen peroxide =  30%

d = Density of hydrogen peroxide =  1.39 g/ml

M = Molar mass  = 34 g/mol

Substituting values;

Co = 10 × 30 ×  1.39/34

Co = 12.26 M

The molarity of the hydrogen peroxide solution from the information supplied in the question is 12.26 M.

Learn more: brainly.com/question/6111443

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The lowest value of the henry's law for methane gas (ch4) will be obtained with __________ as the solvent and a temperature of _
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The lowest value of the henry's law for methane gas (CH₄) will be obtained with H₂O as the solvent and a temperature of  349 K.

The lowest value of the henry's law for methane gas (CH₄) will be obtained with H₂O as the solvent and a temperature of 349 K.  

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How many moles in 2.21 x10E24 atoms of aluminum
lisov135 [29]
<h3>Answer:</h3>

3.67 mol Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.21 × 10²⁴ atoms Al

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2.21 \cdot 10^{24} \ atoms \ Al(\frac{1 \ mol \ Al}{6.022 \cdot 10^{23} \ atoms \ Al})
  2. Divide:                              \displaystyle 3.66988 \ mol \ Al

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.66988 mol Al ≈ 3.67 mol Al

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2 years ago
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